作業:
查詢INFO表所有數據,顯示在頁面上(表格)
性別要顯示男女 民族 顯示民族名稱
1 <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
2 <html xmlns="http://www.w3.org/1999/xhtml">
3 <head>
4 <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
5 <title>無標題文檔</title>
6 </head>
7 <body>
8
9 <table width="1000px" cellpadding="1" border="1"ellspacing="1">
10 <tr>
11 <td>編號</td>
12 <td>姓名</td>
13 <td>性別</td>
14 <td>名族</td>
15 <td>出生日期</td>
16 </tr>
17 <?php 18 //造對象
19 $db=new MySQLi("localhost","root","","hr.sql"); 20 //判斷連接是否出錯
21 !mysqli_connect_error() or die("連接失敗!"); 22 //寫SQL語句
23 $sql="select * from info"; 24
25 //執行SQL
26 $result=$db->query($sql); 27
28 //從結果集對象中讀取數據
29 while($attr=$result->fetch_row()) 30 { 31 $attr[2]=$attr[2]=="1"?"男":"女"; 32
33 $sql="select name from nation where code ='{$attr[3]}'"; 34 $we=$db->query($sql); 35 $mz=$we->fetch_row(); 36
37 echo "<tr> 38 <td>{$attr[0]}</td> 39 <td>{$attr[1]}</td> 40 <td>{$attr[2]}</td> 41 <td>{$mz[0]}</td> 42 <td>{$attr[4]}</td> 43 </tr>"; 44
45
46
47
48 } 49
50
51 ?>
52
53
54
55
56
57
58 </body>
59 </html>
顯示的結果: